\(mdd_{HCl}=100,8.1,19=119,952\left(g\right)\)
\(\Rightarrow m_{HCl}=\dfrac{119,952\times36}{100}=43,18272\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{43,18272}{36,5}=1,18\left(mol\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\left(2\right)\)
Theo PT (1) ta có: \(n_{Zn}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,4.65=26\left(g\right)\)
Theo PT (1): \(n_{HCl\left(PT1\right)}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
\(\Rightarrow n_{HCl\left(PT2\right)}=1,18-0,8=0,38\left(mol\right)\)
Theo PT (2): \(n_{ZnO}=\dfrac{0,38.1}{2}=0,19\left(mol\right)\)
\(\Rightarrow m_{ZnO}=0,19.81=15,39\left(g\right)\)
\(m_{hỗn-hợp-ban-đầu}=26+15,39=41,39\left(g\right)\)
\(\Rightarrow\%_{Zn}=\dfrac{26}{41,39}.100\%=62,81\%\)
\(\Rightarrow\%ZnO=100\%-62,81\%=37,19\%\)